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Got this one right however was wondering if anybody knows how to proceed the algabraic way.. Maybe we can do something like this; x+ x+1 + x+2 + x+3=5K 4x+6=5K ..?

Re: sum of 4 consecutive positive integers [#permalink]
28 Sep 2009, 21:49

1

This post received KUDOS

4X+6=5k

1, x is odd, no hint, wrong

2, none of x-> x+3 is divisible by 5, If not

x is divisible by 5 then 4x+6 divides 5 remains 1 not 0 x+1 is ............5 then sum=4(x+1)+2 divides 5 remains 2 not 0 x+2 is ............5 then sum=4(x+2)-2 divides 5 remains 3 not 0 x+3 is.............5 then sum=4(x+3)-6 divides 5 remains 4 not 0

2 is right

3, 1 Number with the digit is 3, the other should be ended with + 0,1,2 then sum =10k+6 not divisible by 5 + 1,2,4 then sum =10K+10 divisible by 5 these above enough to answer + 2,4,5 then + 4,5,6

Re: sum of 4 consecutive positive integers [#permalink]
29 Sep 2009, 04:14

1

This post received KUDOS

tejal777 wrote:

If the sum of 4 consecutive positive integers is divisible by 5, which of the following must be true?

I. The smallest integer is odd

II. None of the integers is divisible by 5

III. One of the integers has 3 as its units digit

Got this one right however was wondering if anybody knows how to proceed the algabraic way..

I typically think in terms of remainders. If x = 5*k+y (where y = 0,1,2,3 or 4; sometimes I use y = -2,-1,0,1,2). In the above problem x can be replaced with y. it's pretty obvious that the neither of the 4 numbers can be divisible by 5. Otherwise, 4 options -2,-1,0,1 or -1,0,1,2 or 0,1,2,3 (-2) or -3(2), -2,-1,0. clearly, neither of those options work. necessary and sufficient condition is: neither of the 4 numbers is divisible by 5.

I. The smallest integer is odd -> false (6,7,8,9)

II. None of the integers is divisible by 5 -> true (from above)

III. One of the integers has 3 as its units digit -> false (6,7,8,9)

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